مرکزی صفحہ Cracking the SAT Subject Test in Math 2, 2nd Edition: Everything You Need to Help Score a Perfect 800

Cracking the SAT Subject Test in Math 2, 2nd Edition: Everything You Need to Help Score a Perfect 800

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EVERYTHING YOU NEED TO HELP SCORE A PERFECT 800. Equip yourself to ace the SAT Subject Test in Math 2 with The Princeton Review's comprehensive study guide—including 3 full-length practice tests, thorough reviews of key topics, and targeted strategies for every question type.

We don't have to tell you how tough SAT Math is—or how helpful a stellar exam score can be for your chances of getting into your top-choice college. Written by the experts at The Princeton Review, Cracking the SAT Subject Test in Math 2 arms you to take on the test and achieve your highest score.

Techniques That Actually Work.
• Tried-and-true tactics to help you avoid traps and beat the test
• Tips for pacing yourself and guessing logically
• Essential strategies to help you work smarter, not harder

Everything You Need to Know for a High Score.

• Expert subject reviews for every test topic
• Up-to-date information on the SAT Subject Test in Math 2
• Score conversion tables to help you assess your performance and track your progress

Practice Your Way to Perfection.
3 full-length practice tests (2 in the book and 1 online) with detailed answer explanations
• Practice drills throughout each content chapter
• End-of-chapter summaries to help you master key points
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Special thanks to Chris Chimera for his tremendous effort to develop the
new content in this edition. Thanks also to Jonathan Chiu, National ACT
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Title Page

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Part I: Orientation




Part II: Diagnostic Practice Test

Practice Test 1


Practice Test 1: Answers and Explanations
Practice Test 1 Answer Key
Practice Test 1 Explanations
How to Score Practice Test 1

Part III: Content Review

Algebra on the SAT Subject Test in Math 2
Solving Equations
Factoring and Distributing
Plugging In
Plugging In The Answers

Working with Ranges
Direct and Indirect Variation
Work and Travel Questions
Simultaneous Equations
FOIL Method
Factoring Quadratics
The Quadratic Formula
Graphing Calculator to the Rescue!
Comprehensive Algebra Drill

Doing Arithmetic
Fractions, Decimals, and Percentages
Irrational Numbers
Special Exponents
More Important Exponent Stuff
Comprehensive Fundamentals Drill


Plane and Solid Geometry
Plane Geometry Formulas
Rectangular Solids

Tricks of the Trade
Comprehensive Plane and Solid Geometry Drill

Coordinate Geometry
The Coordinate Plane
The Equation of a Line
Linear Inequalities
General Equations
Triaxial Coordinates: Thinking in 3D
Comprehensive Coordinate Geometry Drill


The Basic Functions
Trigonometric Identities
Graphing Trigonometric Functions
Trigonometry in Non-Right Triangles
Polar Coordinates
Comprehensive Trigonometry Drill

10 Functions
Weird Symbols as Functions
Functions Using Standard Notation
Compound Functions

Inverse Functions
Domain and Range
Functions Within Intervals: Domain Meets Range
Graphing Functions
Range and Domain in Graphs
Roots of Functions in Graphs
Symmetry in Functions
Degrees of Functions
Comprehensive Functions Drill
11 Statistics and Sets
Working with Statistics
Permutations, Combinations, and Factorials
Group Questions
Comprehensive Statistics and Sets Drill
12 Miscellaneous
Visual Perception
Arithmetic and Geometric Sequences
Imaginary Numbers
Polynomial Division
Comprehensive Miscellaneous Drill

Part IV: Drills: Answers and Explanations
Chapter 5: Algebra Drill Explanations
Chapter 6: Fundamentals Drill Explanations
Chapter 7: Plane and Solid Geometry Drill Explanations
Chapter 8: Coordinate Geometry Drill Explanations
Chapter 9: Trigonometry Drill Explanations
Chapter 10: Functions Drill Explanations
Chapter 11: Statistics and Sets Drill Explanations
Chapter 12: Miscellaneous Drill Explanations

Part V: Final Practice Test
13 Practice Test 2
14 Practice Test 2: Answers and Explanations
Practice Test 2 Answer Key
Practice Test 2 Explanations
How to Score Practice Test 2

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Part I


Chapter 1
Welcome to the SAT Subject Test in Math 2! This chapter will help you
get familiar with this book and learn how to use it most effectively. We’ll
also talk about when to take the test and how to determine which level to
take. (If you’re flipping through this book in the bookstore, this chapter’s
for you!)

The SAT Subject Test in Math 2 is a standardized test in mathematics.
Colleges use this test to assist in admissions decisions and to place
incoming students in classes at the right level. This test is written by ETS,
a company in the business of writing tests like these. ETS makes money
by charging students to take the SAT and SAT Subject Tests, and charging
again to send the scores to colleges. You’ll also run into ETS exams if you
ever apply to graduate school.
The SAT Subject Test in Math 2 has 50 multiple-choice questions and is
one hour long. The test is scored from 200 to 800 points. The SAT
Subject Test in Math 2 covers a range of mathematical topics, from basic
algebra to trigonometry and statistics.
Many colleges require some SAT Subject Tests (frequently two, but
occasionally one or three). The subjects available are varied: two in
mathematics, three in science, two in history, one in English, and twelve
in foreign languages. Different schools have different preferences and
requirements for which tests to take, too. For example, an engineering
program may want to see one math and one science. Check each school’s
website to determine how many tests you must take and which ones (if
any) are preferred.

What’s on the Test?
The content of the SAT Subject Test in Math 2 is approximately as


Math Level 2


12 questions


10 questions


9 questions

Coordinate Geometry

6 questions

Solid Geometry

3 questions


4 questions


6 questions


50 questions

As you can see, the SAT Subject Test in Math 2 focuses on material you
learned in your Geometry, Algebra II, and Precalculus classes. When it
asks questions about basic concepts, it does so by including the concepts
in a more complicated problem. For example, there are no direct
questions about plane geometry. However, you will need to be able to
apply the concepts of plane geometry to questions about coordinate
geometry or spatial geometry.
You may be overwhelmed by the number of different topics which appear
on the SAT Subject Test in Math 2. Fear not! The test is written with the
expectation that most students have not covered all the material on the
test. Furthermore, you can do well on this test even if you haven’t covered
everything that may show up on the test.

Math Level 1 or 2?
We’d love to say that this decision boils down completely to your comfort
level with the material in each course, but the truth is that not every
school accepts the Math 1 results. You should therefore base your
decision primarily on the admission requirements of the schools that
interest you.

Math 2 is appropriate for high school students who have had a year of
trigonometry or precalculus and have done well in the class. You should
also be comfortable using a scientific or graphing calculator. If you hate
math, do poorly on math tests, or have not yet studied Trigonometry or
Precalculus, the SAT Subject Test in Math 2 is probably not for you. It’s
worth noting, however, that while this test is difficult, the test is scored
on a comparatively generous curve. If you find yourself making random
(or “silly”) mistakes more than anything else, the Math 2 scoring grid
may work in your favor.
Colleges also receive your percentile (comparing you to other test takers),
as well as your scaled (200–800) score. For the most part, they pay
attention to the scaled score and ignore the percentile. However, to the
small extent that percentiles matter, Math 1 has considerably more
forgiving percentiles. People who take Math 2 are generally really good at
math; about 13% of them get a perfect score! Less than 1% of Math 1 testtakers get a perfect score, though. As a result, a 790 on Math 2 is only in
the 85th percentile (about 13% get an 800 and 2% get a 790), while a 790
on Math 1 is still in the 99th percentile. The disparity between the
percentiles continues down the entire score range.
If you are very unsure about which test to take, even after working
practice questions and taking practice tests, you can take both tests.

Want to know which colleges are best for you?
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Advisor app to build your ideal college list and
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The right time to take the SAT Subject Test in Math 2 varies from person
to person. Many students take the test at the end of a Precalculus class in
school. (Precalculus also goes by many other names, such as
Trigonometry, Advanced Functions, or other less recognizable names.)
Some students take Math 2 during or at the end of an AP Calculus course.
The SAT Subject Tests are offered six times per year, and no test date is
easier or harder than any other test date. The most popular test dates are
in May and June, because these are at the end of a school year when the
material is freshest in the student’s mind. Whenever you choose to take
the test, make sure you have time to do some practice beforehand, so that
you can do your best (and not have to take the thing again!).

The Calculator
The SAT Subject Test in Math 2 is designed to be taken with the aid of a
calculator. Students taking this test should have a scientific or graphing
calculator and know how to use it. A “scientific” calculator is one that has
keys for the following functions:
• the values of π and e
• square roots
• raising to an exponent
• sine, cosine, and tangent
• logarithms
Calculators without these functions will not be as useful. Graphing
calculators are allowed. The graphing features on a graphing calculator
are helpful on a fairly small number of questions per test, and they are
necessary for about 0–1 questions per test. If you’re going to take a
graphing calculator to the test, make sure you know how to use it.
Fumbling over your calculator trying to figure something out during the
test is just not a productive use of your time!
This book is going to focus on the TI-84. If you have another family
member of the TI-80 series, know that these comments still apply to you
with minor adjustments. Check with your manual for specific key stroke

changes. If you have a scientific calculator, we’ll be showing you your key
stroke changes in the sidebars throughout the manual.
The ETS Predictor
ETS says that for the SAT Subject Test in Math 2, a calculator may be
useful or necessary for about 55-65 percent of the questions.
Certain kinds of calculators are not allowed. For example, a calculator
with a QWERTY keyboard (like a computer keyboard) is not allowed.
Your calculator must not require a wall outlet for power and must not
make noise or produce paper printouts. There will be no replacements at
the test center for malfunctioning or forgotten calculators, though you’re
welcome to take a spare, as well as spare batteries. Laptops, tablets, and
cell phones are also not allowed as calculators.
Bottom line: You need a calculator for this test. Certain things will be
easier with a graphing calculator, but it is most important that you are
comfortable using your calculator.

It’s best to work through the chapters of this book in sequence, since the
later chapters build on the techniques introduced in earlier chapters. If
you want an overall review of the material on the SAT Subject Test in
Math 2, just start at the beginning and cruise through to the end. This
book will give you all the techniques and knowledge you need to do well
on the test. If you feel a little shaky in certain areas of math and want to
review specific topics, the chapter headings and subheadings will also
allow you to zero in on your own problem topics. As with any subject, pay
particular attention to the math topics you don’t like—otherwise, those
are the ones that will burn you on the real test.

Study Guide
In your free online student tools, you’ll find a
printable copy of two different study guides that
you can use to plot out your progress through
this book.

If you really want to get your money’s worth out of this book, you’ll follow
this study plan.
• Take the diagnostic full-length practice test that appears in Part II.
• Score your test and review it to see where your strengths and
weaknesses lie.
• Focus on the chapters associated with the questions that you got
wrong or didn’t understand.
• Read through each section of a chapter carefully until you feel that you
understand it.
• Try the practice questions at the end of that section.
• Check your answers, and review any questions you got wrong until you
understand your mistakes.
• Once you finish all the sections in a chapter, try the Comprehensive
Drill at the end of that chapter.
• Check your answers, and review any questions you got wrong until you
understand your mistakes.
• Go back and review the chapters which cover material you’re still
struggling with.
• Take the final in-book full-length practice test. Then score and review
If you put in the work and study what’s in this book, you’ll be prepared
for anything that ETS may throw at you on the day of the real test.
Need More?
You can also visit CollegeBoard.org for more
information and test questions. Also, don’t
forget to register your book online for access to
a third full-length practice test!

Chapter 2
It’s easy to get the impression that the only way to excel on the SAT
Subject Test in Math 2 is to become an expert on myriad mathematical
matters. However, there are many effective strategies that you can use.
From Pacing to Process of Elimination to using your calculator, this
chapter takes you through the most important general strategies so you
can start practicing them right way.

It’s true that you have to know some math to do well, but there’s a great
deal you can do to improve your score without staring into math books
until you go blind.
Several important strategies will help you increase your scoring power.
• The questions on the SAT Subject Test in Math 2 are arranged in order
of difficulty. You can think of a test as being divided roughly into
thirds, containing easy, medium, and difficult questions, in that order.
• The SAT Subject Test in Math 2 is a multiple-choice test. That means
that every time you look at a question on the test, the correct answer is
on the paper right in front of you.
• ETS writes incorrect answers on the SAT Subject Test in Math 2 by
studying errors commonly made by students. These are common
errors that you can learn to recognize.
The next few pages will introduce you to test-taking techniques that use
these features of the SAT Subject Test in Math 2 to your advantage, which
will increase your score. These strategies come in two basic types: Section
strategies (which help you determine which questions to do and how
much time to spend on them) and question strategies (which help you
solve an individual question once you’ve chosen to do it.)

The following represents a sample scoring grid from the SAT Subject Test
in Math 2. Note that scoring scales will vary from test to test, so this is
just a general guide.
Raw Score

Scaled Score









































































































































A few things are notable:
• While it’s theoretically possible to score less than a 350, to do so would
require you to score a negative number of raw points (i.e., do worse
than simply randomly guessing). Practically speaking, the scoring

range on the SAT Subject Test in Math 2 is from 350–800.
• On some test dates, some scores are not possible, such as 420 in the
test shown above.
• The scoring grid for the SAT Subject Test in Math 2 is very forgiving,
especially at the top end. Anything from 43 to 50 raw points gets you a
“perfect” 800, and 33 raw points out of a possible 50 is still a 700.
However, the percentiles are brutal: a 700 is only the 61st percentile!

The first step to improving your performance on the SAT Subject Test in
Math 2 is slowing down. That’s right: You’ll score better if you do fewer
questions. It may sound strange, but it works. That’s because the testtaking habits you’ve developed in high school are poorly suited to the SAT
Subject Test in Math 2. It’s a different kind of test.
One Point Over Another?
A hard question on the SAT Subject Test in Math 2 isn’t worth more
points than an easy question. It just takes longer to do, and it’s
harder to get right. It makes no sense to rush through a test if all
that’s waiting for you are tougher and tougher questions—especially
if rushing worsens your performance on the easy questions.
Think about a free-response math test. If you work a question and get the
wrong answer, but you do most of the question right, show your work,
and make a mistake that lots of other students in the class make (so the
grader can easily recognize it), you’ll probably get partial credit. If you do
the same thing on the SAT Subject Test in Math 2, you get one of the four
wrong answers. But you don’t get partial credit for choosing one of the
listed wrong answers; you lose a quarter-point. That’s the opposite of
partial credit! Because the SAT Subject Test in Math 2 gives the opposite
of partial credit, there is a huge premium on accuracy in this test.

How Many Questions Should I Do?
Use the following chart to determine how many questions you should
attempt the next time you take an SAT Subject Test in Math 2:

As you improve, your pacing goals will also get more aggressive. Once you
take your next practice test and score it, come back to this chart and
adjust your pacing accordingly. For example, if you initially scored a 550,
but on your second test you scored a 610, then use the 610–650 line for
your third test, and you may score a 700 (or even higher!).
Your Last Test
For your “last test,” use your last SAT Subject
Test in Math 2 (real or practice), if you’ve taken
one. If you’ve taken the SAT, use your Math
score. You can also use a PSAT score; just add
a “0” to the end of your Math score (so a 56
becomes a 560). If you’ve taken the ACT
instead, multiply your math score by 20 (so a
25 in Math becomes a 500 for the purpose of
pacing on the SAT Subject Test in Math 2). If
you haven’t taken any of these tests, make an
educated guess!

Personal Order of Difficulty
You probably noticed that the previous chart doesn’t tell you which
questions to do on the SAT Subject Test in Math 2, only how many. That’s
because students aren’t all the same. Even if a certain question is easy for
most students, if you don’t know how to do it, it’s hard for you.

Conversely, if a question is hard for most students but you see exactly
how to do it, it’s easy for you. Most of the time, you’ll find lowernumbered questions easy for you and higher-numbered questions harder
for you, but not always, and you should always listen to your Personal
Order of Difficulty (POOD).

Develop a Pacing Plan
The following is an example of an aggressive pacing plan. You should
begin by trying this plan, and then you should adapt it to your own needs.
First, do questions 1–20 in 20 minutes. They are mostly easy, and you
should be able to do each one in about a minute. (As noted above, though,
you must not go so quickly that you sacrifice accuracy.) If there is a
question that looks more time-consuming, but you know how to do it,
mark it so that you can come back to it later, but move on.
Second, pick and choose among questions 21–50. Do only questions that
you are sure you can get right quickly. Mark questions that are more
time-consuming (but you still know how to do them!) so that you can
come back to them later. Cross out questions that you do not know how
to do; you shouldn’t waste any more time on them.
Third, once you’ve seen every question on the test at least once and
gotten all the quick points that you can get, go back to the more timeconsuming questions. Make good choices about which questions to do; at
this point, you will be low on time and need to make realistic decisions
about which questions you will be able to finish and which questions you
should give up for lost.
This pacing plan takes advantage of the test’s built-in order of difficulty
and your POOD. You should move at a brisk but not breakneck pace
through the easy questions so that you have enough time to get them
right but not waste time. You should make sure that you get to the end of
the test and evaluate every question, because you never know if you
happen to know how to do question 50; it may be harder for most
students than question 30, but it just may test a math topic that you

remember very well from class (or this book). Delaying more timeconsuming questions until after you’ve gotten the quick and easy points
maximizes your score and gives you a better sense of how long you have
to complete those longer questions, and, after some practice, it will take
only a few seconds to recognize a time-consuming question.

A Note About Question Numbers
As you cruise through this book, you’ll run into practice questions that
seem to be numbered out of order. That’s because the numbers of the
practice questions tell you what position those questions would occupy on
a 50-question SAT Subject Test in Math 2. The question number gives
you an idea of how difficult ETS considers a given question to be.

It’s true that the math on the SAT Subject Test in Math 2 gets difficult.
But what exactly does that mean? Well, it doesn’t mean that you’ll be
doing 20-step calculations, or huge, crazy exponential expansions that
your calculator can’t handle. Difficult questions on this test require you to
understand some slippery mathematical concepts, and sometimes to
recognize familiar math rules in strange situations.
This means that if you find yourself doing a 20-step calculation, stop.
There’s a shortcut, and it probably involves using one of our techniques.
Find it.
Random Guessing
If you randomly guess on five questions, you
can expect to get one right and four wrong.
Your score for those five questions will be:
This isn’t very helpful.

Process of Elimination (POE)

It’s helpful that the SAT Subject Test in Math 2 contains only multiplechoice questions. After all, this means that eliminating four answers that
cannot possibly be right is just as good as knowing what the right answer
is, and it’s often easier. Eliminating four answers and choosing the fifth is
called the Process of Elimination (POE).
POE Guessing
If you correctly eliminate two answer choices
and guess among the remaining three, you
have a one-in-three chance of getting the right
answer. If you do this on six questions, you can
expect to get two right and four wrong. Your
score will be :
That’s not a lot for six questions, but every
point helps.

POE can also be helpful even when you can’t get down to a single answer.
Because of the way the test is scored (plus one raw point for a correct
answer and minus a quarter-point for an incorrect answer), if you can
eliminate at least one answer, it is to your advantage to guess.
So, the bottom line:
To increase your score on the SAT Subject Test in Math 2, eliminate
wrong answer choices whenever possible, and guess aggressively
whenever you can eliminate anything.
There is a major elimination technique you should rely on as you move
through the test: ballparking.

Sometimes, you can approximate an answer:

You can eliminate answer choices by ballparking whenever you have
a general idea of the correct answer. Answer choices that aren’t even
in the right ballpark can be crossed out.
Take a look at the following three questions. In each question, at least one
answer choice can be eliminated by ballparking. See whether you can
make eliminations yourself. For now, don’t worry about how to do these
questions—just concentrate on eliminating answer choices.

6. If

= 1.84, then x2 =

(A) −10.40
(B) −3.74
(C) 7.63
(D) 10.40
(E) 21.15
Here’s How to Crack It
You may not have been sure how to work with that ugly fractional
exponent. But if you realized that x2 can’t be negative, no matter what x
is, then you could eliminate (A) and (B)—the negative answers—and then
guess from the remaining answer choices.

Figure 1

13. In Figure 1, if c = 7 and θ = 42°, what is the value of a ?
(A) 0.3
(B) 1.2
(C) 4.7
(D) 5.2
(E) 6.0
Here’s How to Crack It
Unless you’re told otherwise, the figures that the SAT Subject Test in
Math 2 gives you are drawn accurately, and you can use them to ballpark.
In this example, even if you weren’t sure how to apply trigonometric
functions to the triangle, you could still approximate based on the
diagram provided. If c is 7, then a looks like, say, 5. That’s not specific
enough to let you decide between (C), (D), and (E), but you can eliminate
(A) and (B). They’re not even close to 5. At the very least, that gets you
down to a 1-in-3 guess—much better odds.
Can I Trust The Figure?
In order to intentionally mislead you,
sometimes ETS inserts figures that are
deliberately inaccurate. When the figure is
wrong, ETS will print underneath, “Note: Figure
not drawn to scale.” When you see this note,
trust the text of the problem, but don’t believe
the figure, because the figure is just there to
trick you.

22. The average (arithmetic mean) cost of Simon’s math
textbooks was $55.00, and the average cost of his history
textbooks was $65.00. If Simon bought 3 math
textbooks and 2 history textbooks, what was the average

cost of the 5 textbooks?
(A) $57.00
(B) $59.00
(C) $60.00
(D) $63.50
(E) $67.00
Here’s How to Crack It
Here, once again, you might not be sure how to relate all those averages.
However, you could realize that the average value of a group can’t be
bigger than the value of the biggest member of the group, so you could
eliminate (E). You might also realize that, since there are more $55 books
than $65 books, the average must be closer to $55.00 than to $65.00, so
you could eliminate (C) and (D). That gets you down to only two answer
choices, a 50/50 chance. Those are excellent odds.

These are all fairly basic questions. By the time you’ve finished this book,
you won’t need to rely on ballparking to answer them. The technique of
ballparking will still work for you, however, whenever you’re looking for
an answer you can’t figure out with actual math.

“Better” Than Average
What makes a question hard? Sometimes, a hard question tests more
advanced material. For example, on the SAT Subject Test in Math 2,
questions about polar coordinates are rare before question 20.
Sometimes a hard question requires more steps, four or five rather than
one or two. But more often, a hard question has trickier wording and
better trap answers than an easy question.
ETS designs its test around certain trends and traps, looking to catch the
average student with the sort of tricks and problems that have tripped
test-takers up in the past. While this does mean that you’ll have to be
alert, it also means that many of these questions have predictable wrong
answers, and you can use this knowledge to “beat” the curve. When ETS

writes a question that mentions “a number,” it counts on students to
think of numbers like 2 or 3, not numbers like −44.76 or 4π. ETS counts
on students to think of the most obvious thing, like 2 or 3 instead of
−44.76 or 4π. Don’t be led astray by the urge to choose these; instead, use
it to your advantage.
There is nothing on the SAT Subject Test in Math 2 that hasn’t been
taught to students, which means that in order to trip students up, the test
writers need to make students pick a wrong answer. It does that by
offering answers that are too good to be true: tempting
oversimplifications, obvious answers to subtle questions, and all sorts of
other answers that seem comforting and familiar. Take a step back. Try
eliminating choices like these and then pick and check another one

28. Ramona cycles from her house to school at 15 miles per
hour. Upon arriving, she realizes that it is Saturday and
immediately cycles home at 25 miles per hour. If the
entire round-trip takes her 32 minutes, then what is her
average speed, in miles per hour, for the entire roundtrip?
(A) 17.0
(B) 18.75
(C) 20.0
(D) 21.25
(E) 22.0
Here’s How to Crack It
This is a tricky problem, and you may not be sure how to solve it. You
can, however, see that there’s a very tempting answer among the answer
choices. If someone goes somewhere at 15 mph and returns at 25 mph,
then it seems reasonable that the average speed for the trip should be 20
mph. For question 28, however, that’s far too obvious to be right.
Eliminate (C).

Stop and Think
Anytime you find an answer choice immediately appealing on a hard
question, stop and think again. ETS collects data from thousands of
students in trial tests before making a question a scored part of their
tests. If it looks that good to you, it probably looked good to many of
the students taking the trial tests. That attractive answer choice is
almost certainly a trap. The right answer won’t be the answer most
people would pick. On hard questions, obvious answers are wrong.
Eliminate them.

34. If θ represents an angle such that sin2θ = tanθ − cos2θ,
then sinθ − cosθ =
(A) −
(B) 0
(C) 1
(D) 2
(E) It cannot be determined from the information
Here’s How to Crack It
On a question like this one, you might have no idea how to go about
finding the answer. That “It cannot be determined” answer choice may
look awfully tempting. You can be sure, however, that (E) will look
tempting to many students. It’s too tempting to be right on a question
this hard. You can eliminate (E).

48. If the above cones are similar, and the volume of the
larger cone is 64, then what is the volume of the smaller
(A) 2
(B) 4
(C) 8
(D) 16
(E) 32
Here’s How to Crack It
This one may seem simple: the smaller cone is half as tall as the larger
cone, so its volume must be

= 32. But wait! This is question number

48. That means that most test takers will miss it. We’ll cover how to
tackle this question easily in the Solid Geometry chapter, but before you
turn the page, be sure to cross out 32, as it's a trap answer!


The techniques in this book will go a long way toward increasing your
score, but there’s a certain minimum amount of mathematical knowledge
you’ll need in order to do well on the SAT Subject Test in Math 2. We’ve
collected the most important rules and formulas into lists. As you move
through the book, you’ll find these lists at the end of each chapter.
The strategies in this chapter, and the techniques in the rest of this book,
are powerful tools. They will make you a better test taker and improve
your performance. Nevertheless, memorizing the formulas on our lists is
as important as learning techniques. Memorize those rules and formulas,
and make sure you understand them.

Using That Calculator
Behold the First Rule of Intelligent Calculator Use:
Your calculator is only as smart as you are.
It’s worth remembering. Some test takers have a dangerous tendency to
rely too much on their calculators. They try to use them on every question
and start punching numbers in even before they’ve finished reading a
question. That’s a good way to make a question take twice as long as it
has to.
The most important part of problem solving is done in your head. You
need to read a question, decide which techniques will be helpful in
answering it, and set up the question. Using a calculator before you really
need to do so will keep you from seeing the shortcut solution to a
Scientific or Graphing?
ETS says that the SAT Subject Test in Math 2
is designed with the assumption that most test
takers have graphing calculators. ETS also
says that a graphing calculator may give you
an advantage on a handful of questions. If you

have access to a graphing calculator and know
how to use it, you may want to choose it
instead of a scientific calculator.

When you do use your calculator, follow these simple procedures to avoid
the most common calculator errors.
• Check your calculator’s operating manual to make sure that you know
how to use all of your calculator’s scientific functions (such as the
exponent and trigonometric functions).
• Clear the calculator at the beginning of each problem to make sure it’s
not still holding information from a previous calculation.
• Whenever possible, do long calculations one step at a time. It makes
errors easier to catch.
• Write out your work! Label everything, and write down the steps in
your solution after each calculation. That way, if you get stuck, you
won’t need to do the entire problem over again. Writing things down
will also prevent you from making careless errors.
• Keep an eye on the answer choices to see if ETS has included a partial
answer designed to tempt you away from the final answer. Eliminate
Above all, remember that your brain is your main problem-solving tool.
Your calculator is useful only when you’ve figured out exactly what you
need to do to solve a problem.

Set It Up!
Some questions on the SAT Subject Test in
Math 2 can be answered without much
calculation—the setup itself makes the answer
clear. Remember: Figure out how to do the
problem with your brain; then do the problem
with your calculator.

Part II
Diagnostic Practice Test

Practice Test 1
Practice Test 1: Answers and Explanations

Chapter 3
Practice Test 1
Click here to download and print the PDF version of this exercise.

For each of the following problems, decide which is the BEST of the
choices given. If the exact numerical value is not one of the choices, select
the choice that best approximates this value. Then fill in the
corresponding oval on the answer sheet.
Notes: (1) A scientific or graphing calculator will be necessary for
answering some (but not all) of the questions on this test. For each
question, you will have to decide whether or not you should use a
(2) The only angle measure used on this test is degree measure. Make
sure that your calculator is in degree mode.
(3) Figures that accompany problems on this test are intended to
provide information useful in solving the problems. They are drawn as
accurately as possible EXCEPT when it is stated in a specific problem that
its figure is not drawn to scale. All figures lie in a plane unless otherwise
(4) Unless otherwise specified, the domain of any function f is assumed
to be the set of all real numbers x for which f(x) is a real number. The
range of f is assumed to be the set of all real numbers f(x), where x is in
the domain of f.
(5) Reference information that may be useful in answering the
questions on this test can be found below.

Volume of a right circular cone with radius r and height h: V = πr2h
Lateral area of a right circular cone with circumference of the base c
and slant height ℓ: S = cℓ
Volume of a sphere with radius r: V = πr3
Surface area of a sphere with radius r: S = 4πr2
Volume of a pyramid with base area B and height h: V = Bh

1. If 2y + 6 =

(y + 3) for all y, then c =

(B) 2
(C) 9
(D) 15
(E) 18
2. The relationship between a temperature F in degrees Fahrenheit and
a temperature C in degrees Celsius is defined by the equation F = C
+ 32, and the relationship between a temperature in degrees

Fahrenheit and a temperature R in degrees Rankine is defined by
the equation R = F + 460. Which of the following expresses the
relationship between temperatures in degree Rankine and degrees
(A) R = C − 32 + 460
(B) R = C + 32 + 460
(C) R = C + 32 − 460
(D) R = C + 860
(E) R = C − 828
3. What is the slope of a line containing the points (1, 13) and (−3, 6)?
(A) 0.14
(B) 0.57
(C) 1.75
(D) 1.83
(E) 6
4. If a + b + c = 12, a +b = 4, and a + c = 7, what is the value of a ?
(A) 2
(B) 1
(D) 2

5. If g(x) = 2ex − 2 and h(x) = ln(x), then g(h(7)) =
(A) 7.69
(B) 12
(C) 14
(D) 26.43
(E) 31.98
6. The intersection of a cylinder and a plane could be which of the
I. A circle
II. A triangle
III. A rectangle
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II, and III

7. The figure above shows a helium balloon rising vertically. When the

balloon reaches a height of 54 inches, the angles of elevation from
points X and Y on the ground are 72.4° and 50.8°, respectively.
What is the distance, in inches, between points X and Y ?
(A) 61.17
(B) 72.29
(C) 84.15
(D) 124.72
(E) 236.44
8. What is the value of y2 if


(A) 2562
(B) 256
(C) 16
(D) 4
(E) 2
9. The points in the xy-plane are transformed so that each point A(x, y)
is transformed to A’ (3x, 3y). If the distance between point A and the
origin is c, then the distance between the point A’ and the origin is
(C) c
(D) c
(E) 3c

10. If
(A) x2 − 2


, then q(x) =

(B) x2
(C) x
11. If x is the degree measure of an angle such that 0° < x < 90° and
cosx = 0.6, then sin (90° − x) =
(A) 0.4
(B) 0.5
(C) 0.6
(D) 0.7
(E) 0.8
12. The set of points defined by the equation x2 + y2 + z2 = 4 is
(A) a point
(B) a line
(C) a circle
(D) a plane
(E) a sphere
13. The graph of the function g, where
vertical asymptote at x =
(A) 0 only
(B) 3 only
(C) 7 only
(D) 0 and 3 only
(E) 0, 3, and 7

, has a

14. The graph of y = x4 + 8x3 − 4x2 − 64x + k is shown above. Which of
the following could be the value of k?
(A) 1,240
(E) −1,240
15. If sinx = 0.6743, then cscx =
(A) 0.6481
(B) 0.8374
(C) 1.2953
(D) 1.4830
(E) 1.9637
16. Sarah is planning a vacation at a hotel that costs $80 per night.
Sarah must also pay the $170 airfare to get there and will also pay
for an equally priced hotel room for a friend who will be visiting her
on three of the nights. Which of the following correctly expresses the
average cost, in dollars, for each night as a function of n, the number
of nights of the vacation?

17. Which of the following is an equation whose graph is a set of points
equidistant from the points (0, 0) and (6, 0)?
(A) x = 3
(B) y = 3
(C) x = 3y
(D) y = 3x
(E) y = 3x + 3
18. What is the sum of the infinite geometric series

(D) 1

19. Which of the following is equivalent to a− b ≥ a + b?
(A) a ≤ b
(B) a ≤ 0
(C) b ≤ a
(D) b ≤ 0
(E) b ≥ 0
20. If m and n are in the domain of a function g and g(m) > g(n), which
of the following must be true?
(A) mn ≠ 0
(B) m > n
(C) m < n
(D) m = n
(E) m ≠ n
21. In a certain office, the human resources department reports that
60% of the employees in the office commute over an hour on
average each day, and that 25% of those employees who commute
over an hour on average each day commute by train. If an employee
at the office is selected at random, what is the probability that the
employee commutes over an hour on average by train?
(A) 0.10
(B) 0.15
(C) 0.20
(D) 0.25
(E) 0.30
22. To the nearest degree, what is the measure of the second smallest
angle in a right triangle with sides 5, 12, and 13 ?

(A) 23
(B) 45
(C) 47
(D) 60
(E) 67
23. Which of the following is an equation of a line perpendicular to y =
3x − 5 ?
(A) y = 5x − 3
(B) y = −3x + 5
(C) y = x + 5
(D) y = − x + 4
24. What is the range of the function g(x) = −2 + 5cos (3x + 7π) ?
(A) −1 ≤ g(x) ≤ 1
(B) −5 ≤ g(x) ≤ −1
(C) −5 ≤ g(x) ≤ 5
(D) −7 ≤ g(x) ≤ 3
(E) −7 ≤ g(x) ≤ 5
25. Of the following list of numbers, which has the greatest standard
(A) 1, 2, 3
(B) 2, 2, 2
(C) 2, 4, 6
(D) 4, 7, 10
(E) 6, 8, 10

26. The formula F =Ie0.06y gives the final amount F that a bank account
will contain if an initial investment I is compounded continuously at
an annual interest of 6% for y years. Using this formula, after how
many years will an initial investment of $100 be worth
approximately $600?
(A) 5.2
(B) 6.0
(C) 13.0
(D) 22.4
(E) 29.7

27. If cosθ < 0 and

> 0, then θ must be in which quadrant in the

figure above?
(A) I
(B) II
(D) IV
(E) There is no quadrant in which both conditions are true.
28. If g(−x) = −g(x) for all real numbers x and if (4, 9) is a point on the
graph of g, which of the following points must also be on the graph

of g ?
(A) (−9, −4)
(B) (−4, −9)
(C) (−4, 9)
(D) (4, −9)
(E) (9, 4)
If a is a multiple of 10, then a is a multiple of 5.
29. If a is an integer, which of the following CANNOT be inferred from
the statement above?
(A) If a is a multiple of 5, then a is a multiple of 10.
(B) If a is not a multiple of 5, then a is not a multiple of 10.
(C) a is a multiple of 10 implies that a is a multiple of 5.
(D) A necessary condition for a to be a multiple of 10 is that a is a
multiple of 5.
(E) In order for a to be a multiple of 5, it is sufficient that a be a
multiple of 10.
30. In how many different orders can 8 different colors of flowers be
arranged in a straight line?
(E) 16,777,216
31. What value does
(A) 0
(B) 0.5
(C) 1

approach as x approaches 0 ?

(D) 2
(E) It does not approach a unique value
32. If f(x) = |7 − 5x|, then f(1) =
(A) f(1)
(B) f(0)
(D) f(2)
33. What is the period of the graph of y = 3tan (2πx + 9) ?
(C) 3

34. The figure above shows a map of Maple Street and Elm Street.
Katherine is biking from Point X to Point Y. The straight-line
distance from Point X to Point Y is 40 kilometers. If Katherine bikes
at an average speed of 15 km per hour along Maple Street and Elm
Street, how long will it take Katherine to get to Point Y ?
(A) 40 minutes
(B) 2 hours and 35 minutes
(C) 2 hours and 40 minutes
(D) 3 hours and 15 minutes
(E) 3 hours and 35 minutes












35. If g is a polynomial of degree 4, five of whose values are shown in
the table above, then g(x) could equal

(x + 1)(x + 2)2

(A) g(x) =

(B) g(x) = (x − 2)(x − 1)(x + 2)(x + 3)
(C) g(x) = (x − 2)

(x + 1)(x + 2)

(D) g(x) = (x − 3)(x − 2)(x − 1)(x + 2)
(E) g(x) = (x − 2)(x − 1)

(x + 2)

36. The only prime factors of an integer m are 2, 3, 5, and 13. Which of
the following could NOT be a factor of m ?
(A) 6
(B) 9
(C) 12
(D) 26
(E) 35
37. If 0 ≤ x ≤

and cosx = 4sinx, what is the value of x ?

(A) 0.245
(B) 0.250
(C) 0.328
(D) 1.217
(E) 1.326
38. If
(A) 0.04
(B) 1.73
(C) 3.17
(D) 5.00
(E) 25.98

, what is the value of g−1(15) ?

39. The Triangular Number Sequence Tn can be defined recursively as
T1 = 1
Tn = Tn − 1 + n for n > 1
What is the 11th term of the sequence?
(A) 45
(B) 55
(C) 66
(D) 78
(E) 91
40. If f(x) = x3 + x2 − 16x + 12, which of the following statements are
I. The equation f(x) = 0 has three real solutions
II. f(x) ≥ −8 for all x ≥ 0
III. The function is increasing for x > 2
(A) I only
(B) III only
(C) I and III only
(D) II and III only
(E) I, II, and III only

41. Portions of the graphs of g and h are shown above. Which of the
following could be a portion of the graph of gh ?





42. The set of all real numbers y such that


(A) all real numbers
(B) no real numbers
(C) negative real numbers only
(D) nonnegative real numbers only
(E) zero only

43. In the triangle shown above, sinx =
44. The length, width, and height of a rectangular solid are 6, 3, and 2.
What is the length of the longest segment that can be drawn
between two vertices of the solid?
(A) 6
(B) 3
(C) 7

(D) 12
(E) 18
45. If logn2 = a and logn5 = b, then logn50 =
(A) a + b
(B) a + b2
(C) ab2
(D) a + 2b
(E) a + 5b
46. If cosx = a, then, for all x, in the interval 0 < x <

, tanx =

(A) a2 + 1


47. Which of the following shifts in the graph of y = x2 would result in
the graph of y = x2 + 4x + c, where c is a constant greater than 5?
(A) Left 2 units and up c − 4 units
(B) Right 2 units and down c − 4 units
(C) Right 2 units and down c + 4 units
(D) Left 2 units and up c + 4 units
(E) Right 4 units and up c units

48. If the height of a right square pyramid is increased by 12%, by what
percent must the side of the base be increased, so that the volume of
the pyramid is increased by 28%?
(A) 3%
(B) 7%
(C) 10%
(D) 36%
(E) 56%
49. If Matrix X has dimensions a ×b and Matrix Y has dimensions b × c,
where a, b, and c are distinct positive integers, which of the
following must be true?
I. The product XY exists and has dimensions a × c.
II. The product XY exists and has dimensions b × b.
III. The product YX does not exist.
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) II and III only

50. If z is the complex number shown in the figure above, which of the
following could be iz?
(A) A
(B) B
(C) C
(D) D
(E) E


Chapter 4
Practice Test 1: Answers and


1. E

The question asks for the value of c in an equation that is true
for all values of y, so plug in a value for y. To make the math
easy on the right side, plug in y = 6. Substitute y = 6 into the
equation to get 2(6) + 6 =

(6 + 3). Simplify to get 18 =


Cancel the 9s on the right side to get 18 = c. The correct answer
is (E).
2. B

The question asks for the relationship between R and C. Since
there are variables in the choices, Plug In. Let C = 10. If C = 10,
then F = (10) + 32 = 50. If F = 50, then R = 50 + 460 = 510.
Plug C = 10 and R = 510 into each choice and eliminate any that
aren’t true. Choice (A) is 510 =

(10) − 32 + 460. This is false,

so eliminate (A). Choice (B) is 510 = (10) + 32 + 460. This is
true, so keep (B). Choice (C) is 510 =

(10) + 32 − 460. This is

false, so eliminate (C). Choice (D) is 510 =

(10) + 860. This is

false, so eliminate (D). Choice (E) is 510 = (10) − 828. This is
false, so eliminate (E). The correct answer is (B).

3. C

To find the slope of a line, use the slope formula: slope =
. Let (x1, y1) = (1, 13) and (x2, y2) = (−3, 6). The slope of
the line is

4. B

. The correct answer is (C).

The question includes three equations and three variables, so
find a way to combine the equations. Because the second
equation provides a value for a + b, substitute this value into
the equation for a + b + c to get the value of c. When a + b = 4 is
substituted into a + b +c = 12, the result is 4 + c = 12. Subtract 4
from both sides to get c = 8. Now, substitute c = 8 into a + c = 7
to get a + 8 = 7. Subtract 8 from both sides to get a = −1. The
correct answer is (B).

5. B

The question asks for g(h(7)). On questions involving
composition of functions, start on the inside and work toward
the outside. Find h(7). Since h(x) = ln(x), h(7) = ln(7) ≈
1.94591015. To find g(h(7)), find g(1.94591015). Since g(x) =
2ex − 2, g(1.94591015) = 2e1.94591015 − 2 = 12. The correct
answer is (B).

6. C

The question asks for which of the three listed figures could be
the intersection of a plane and a cylinder. Go through one at a
time. For (I), since the bases of a cylinder are circles, the plane
could intersect the cylinder in a way that the plane contains one
of the bases and forms a circle. Since the intersection could
form a circle, (I) must be included. Eliminate (B) and (D),
which don’t include (I). Try (II). There doesn’t seem to be an
obvious way to form a triangle. However, don’t eliminate (II)
right away in case there is a way that isn’t obvious. Try (III).

Determine whether a rectangle can be formed. If the plane
passes through the diameters of each base, then a rectangle is
formed. Therefore, (III) must be included, so eliminate (A).
Now, come back to (II). If the plane is parallel to the bases, a
circle is formed rather than a triangle. If the plane is
perpendicular to the bases, a rectangle is formed. If the plane is
at any other angle, a curved path is formed, so the result cannot
be a triangle. Therefore, eliminate (E), which includes (II). The
correct answer is (C).
7. A

The question asks for the distance between X and Y. Use the
vertical height to form two right triangles. Find the base of each
triangle. The sum of the two bases will be the distance between
X and Y. For reference, call the balloon point Z and the point on
the ground directly below the balloon point W. Look at triangle
XWZ. Angle X is 72.4°. WZ, which is opposite the angle X, is 54.
The needed side is XW, which is adjacent to the angle X.
Therefore, tan

. Multiply both sides by a to get

atan (72.4°) = 52. Divide both sides by tan (72.4°) to get
. Now, do the same for triangle YWZ.
Angle Y is 50.8°. WZ, which is opposite angle Y, is 54. The
needed side is YW, which is adjacent to angle Y. Therefore,


. Multiply both sides by a to get atan (50.8°)




tan (50.8°)



. Add XW to WZ to get 17.13 + 44.04 =

61.17. The correct answer is (A).
8. B

The question asks for the value of y2. Since
square both sides to get y2 = 342 − 302. Put the right side of the
equation into a calculator to get 342 − 302 = 1,156 − 900 = 256.
The correct answer is (B).

9. E

Because there are variables in the choices, plug in. Pick
coordinates for point A. Because the question involves distance,
choose a point that can be used to make a Pythagorean triple.
Let A be (x, y) = (3, 4). Point A’ is (3x, 3y) = (9, 12). The
distance between A and the origin is c. Draw a segment
vertically from A to the x-axis, forming a right triangle. The
distance to the x-axis is 4, and the distance along the x-axis is 3.
Therefore, this is a 3:4:5 right triangle, and c = 5. Do the same
for A’. Draw a vertical line from (9, 12), forming a right triangle.
The horizontal side has a length of 9, and the vertical side has a
length of 12. Therefore, this is a 9:12:15 right triangle, and the
distance from A’ to the origin is 15, which is the target.
(Alternatively, use the Pythagorean Theorem to determine the
hypotenuse of both triangles.) Plug c = 5 into each choice and
eliminate any that aren’t 15. Choice (A) is
Choice (B) is

, so eliminate (A).

, so eliminate (B). Choice (C) is 5, soeliminate

(C). Choice (D) is 5

, so eliminate (D). Choice (E) is 3(5) = 15,

so keep (E). The correct answer is (E).
10. D

There are variables in the choices, so plug in. Let x = 3. Then,
≈ 15.389. This is the target.
Go through the choices, one at a time and determine for which
expression for q(x) does p(q(3)) ≈ 15.389. For (A), if q(3) = 32 −
2 = 7, then

. Eliminate (A). For

(B), if q(3) = 32 = 9, then









. Eliminate (C). For (D), if
, then










so eliminate (E).
The correct answer is (D).
11. C

The question asks for sin (90° − x). There are two possible
approaches to this problem. One is to find the value of x by
using the inverse cosine function. If cosx = 0.6, then take the
inverse cosine of both sides to get x = cos−1(0.6) ≈ 53.13.
Therefore, sin (90° − x) ≈ sin (90° − 53.13°) ≈ 0.6. Alternatively,
use the identity cosx = sin (90° − x). Therefore, if cosx = 0.6,
then sin (90° − x) = 0.6 Using either method, the correct
answer is (C).

12. E

In xyz-coordinates an equation with the graph x2 + y2 + z2 = r2
is a sphere with radius r and center at the origin. However, if
this equation is not familiar, the question can still be answered
using POE. Find points that satisfy this equation. Start with
points (2, 0, 0), (0, 2, 0), and (0, 0, 2). Because there is more
than one point, eliminate (A). These three points do not form a
line, so eliminate (B). These points could make a circle, plane,
or sphere, so Plug In more points. Try (−2, 0, 0), (0, −2, 0), and
(0, 0, −2). These six points are not on the same plane, so
eliminate (D). Since all points in any circle must be on the same
plane, eliminate (C), as well. Only one choice remains. The
correct answer is (E).

13. B

The question asks for the x-values at which g has vertical
asymptotes. A function has a vertical asymptote for x-values at
which the denominator is 0 and the factor that makes the
denominator equal to 0 cannot be canceled out with the
numerator. Since the numerator of g cannot be factored, only
worry about where the denominator is 0. Set x2 − 6x + 9 = 0.
Factor the left side, finding two factors of 9 with a sum of −6.
These are −3 and −3. Therefore, the factored form of the
equation is (x − 3)(x − 3) = 0. Set both factors to 0 and solve. In
both cases, the equation is x − 3 = 0, so add to both sides to get
x = 3. The correct answer is (B).

14. C

The question asks for the value of k, which is the constant term
in the polynomial. The constant term represents the yintercept. According to the graph, the curve crosses the y-axis
between 0 and 40. Only one choice is between these. The
correct answer is (C).

15. C

The question asks for cscx, which is equivalent to


Substitute the value of sinx given by the question to get csc






16. D

The question asks for the average cost for each night, which is
. There are variables in the choices, so plug in. Let n
= 4. Since n represents the number of nights, let this be the
denominator. Determine the total cost. Her stay at the hotel
costs $80 per night for four nights for a total of 4 × $80 = $320.
Furthermore, the three-night hotel stay for her friend costs 3 ×
$80 = $240. She must also pay airfare, which is $170.
Therefore, the total cost is $320 + $240 + $170 = $730, and the
average cost is

. This is the target number. Plug n

= 4 into each of the choices, and eliminate any that aren’t 182.5.
Choice (A) is

. Eliminate (A). Choice (B) is















. Eliminate (E). The correct answer is (D).
17. A

The set of points equidistant between two points is the
perpendicular bisector of the segment whose endpoints are the
two points. The segment with endpoints (0, 0) and (6, 0) lies on
the line y = 0 and has midpoint (3, 0). Since a line in the form y
= c, where c is a constant, is a horizontal line, the perpendicular
line must be a perpendicular line the form x = k, where k is a
constant. To be a bisector, the line must go through the
midpoint, which is (3, 0), so the line is x = 3. Alternatively,
sketch the two points and sketch each of the choices. Choices
(B), (C), (D), and (E) all have points that are clearly closer to (0,
0) to (6, 0) and vice versa, so they can be eliminated. The
correct answer is (A).

18. B

A geometric series is one with nth term arn−1, where a
represents the first term, and r represents the common ratio,
i.e. the number by which each term must be multiplied to get
the next term. If 0 < r < 1, then the sum of an infinite series can
be found using the formula

. The first term is , fill in this

for a. To find the common ratio, set up the equation
Multiply both sides by 9 to get


. Therefore, the sum


. Alternatively, find the sum of the

known terms on a calculator
This is close to


Using either method, the correct

answer is (B).
19. D

Simplify the inequality by combining like terms. Subtract a
from both sides to get −b ≥ b. Add b to both sides to get 0 ≥ 2b.
Divide both sides by 2 to get 0 ≥ b. This can be rewritten as b ≤
0. The correct answer is (D).

20. E

The question says must be true, so eliminate any choice that can
be false. It is unknown whether g is increasing, decreasing, or
neither, so attempt as many cases as is needed to eliminate four
choices. Let g be an increasing function. Try g(x) = x. In this
case, if g(m) >g(n), then m > n. Eliminate (C) and (D).
Furthermore, it could be that m = 2 and n = 0. In this case, mn
= 0, so eliminate (A) as well. Now try g(x) = −x. In this case, if
g(m) > g(n), −m > −n. Divide both sides by −1 to get m <n.
Eliminate (B). The correct answer is (E).

21. B

The question involves percents, so plug in 100 as the total
number of employees in the office. 60% of the employees
commute over an hour each day on average. Since 60% of 100 is
60, exactly 60 employees commute over an hour each day on
average. 25% of those 60 employees commute by train. Since
25% of 60 is

, there are 15

employees who commute over an hour by train. The question
asks for the probability that a randomly selected employee
commutes over an hour on average by train. There are 15
employees who do this out of 100 total employees, so the
probability is
22. E

= 0.15. The correct answer is (B).

The second smallest angle is the one opposite the second
shortest side. The second shortest side is the 12. Mark the angle
opposite the 12 as x°. The hypotenuse is 13. Therefore, sin
. Since sinx =

take sin−1 of both sides to get x ≈

67.38. The question asks for the nearest degree, so the correct
answer is (E).
23. D

The question asks for the equation of the line perpendicular to y
= 3x − 5. In the xy-plane, perpendicular lines have slopes that
are negative reciprocals. Find the slope of y = 3x − 5. For a line
in the form y = mx + b, the slope is represented by m.
Therefore, the slope of y = 3x − 5 must be 3. Find a choice that
represents the equation of a line with a slope of − . Choice (E)
is not the equation of a line, so eliminate (E). The remaining
choices are in y = mx + b form, so select the choice with m = −
. The correct answer is (D).

24. D

The question asks for the range of g(x), which is a cosine
function. The range of cos (x) is −1 ≤ cos (x) ≤ 1. Anything that’s
inside the parentheses does not affect the range. The coefficient
3 only affects the period and not the range. Similarly, 7π only
shifts the graph to the left and does not affect the range.
Therefore, the range of cos (3x + 7π) is −1 ≤ cos (3x + 7π) ≤ 1.
The coefficient 5 changes the amplitude, which is defined as
half the distance from the maximum and minimum values of
the cosine graph. Therefore, the range of 5cos (3x + 7π) is −5 ≤
5cos (3x + 7π) ≤ 5. The constant −2 represents a downward
shift, so subtract 2 from both the maximum and the minimum.
Therefore, the range of g(x) is −7 ≤ g(x) ≤ 3. The correct answer
is (D).

25. D

The question asks for the list with the greatest standard
deviation. The list with the greatest standard deviation is the list
in which the numbers are farthest apart. The numbers in (D)
have the greatest separation. The correct answer is (D).

26. E

First, plug in the known values into the equation. The initial
investment is $100, so plug in I = 100. The account needs to be
worth $600, so plug in F = 600 to get 600 = 100e0.06y. Solve for
y. Divide both sides by 100 to get 6 = e0.06y. Take the natural
log of both sides to get ln (6) = 0.06y and 1.791759 = 0.06y.
Divide both sides by 0.06 to get 29.86 = y, which is closest to
29.7. The correct answer is (E).

27. C

The question asks for the location of θ on the xy-plane. Use the
unit circle. In the unit circle, cosθ is equivalent to the xcoordinate, and sinθ is equivalent to the y coordinate. If cosθ <
0, then x < 0. The x-coordinate is negative in quadrants II and
III. Eliminate (A) and (D). The question also states that
. If a fraction is positive, then the numerator and

denominator must be of the same sign. Since cosθ is negative,
so is sinθ. Thus, the y-coordinate must be negative. The x- and
y-coordinate are both negative in Quadrant III, so the correct
answer is (C).
28. B

The question asks for what point must be on the graph of g. The
only point on the graph of g that is given is (4, 9). Therefore,
g(4) = 9. Since g(−x) = −g(x), g(−4) = −g(4) = −9. Since g(−4)
= −9, the point (−4, −9) must also be on the graph of g. The
correct answer is (B).

29. A

The question asks for what statement CANNOT be inferred
from the given statement. Because the question says CANNOT,
ignore the CANNOT and mark each choice as Y or N depending
of whether it can be inferred. The given statement is a
conditional statement in the form if p then q. In the original
statement, p is a multiple of 10 and q is a multiple of 5. A
conditional statement is logically equivalent to its
contrapositive, which is in the form if not q, then not p.
Therefore, the contrapositive of the given statement If a is a
multiple of 10, then a is a multiple of 5 is If a is not a multiple
of 5, then a is not a multiple of 10. Go through each choice one
at a time. Choice (A) is in the form if q then p. A reversal of the
order of the original condition statement cannot be assumed to
be equivalent to the original. (For example, consider a = 15.)
Mark this choice as N. Choice (B) is the contrapositive. The
contrapositive is logically equivalent, so mark this choice with
Y. Choice (C) rephrases an if p then q statements as an p implies
q. This is logically equivalent, so mark this choice with Y. Choice
(D) discusses a necessary condition. For a statement in the form
if p then q, p is referred to as the sufficient condition and q is
referred to as the necessary condition. Since (D) refers to q as
the necessary condition, mark (D) with Y. Similarly, (E) refers
to p as the sufficient condition, so mark (E) with Y. Four choices
are marked as Y and one choice is marked with N, so select the

choice marked with N. The correct answer is (A).
30. C

The question asks for how many different arrangements of 8
flowers in a line. Draw 8 spaces in a line for the 8 flowers.
Consider the number of possibilities for each space. For the first
space, any of the 8 flowers can be chosen, so put an 8 in the first
space. Once that flower is chosen, there are 7 possible flowers
remaining for the second choice, so put a 7 in the second space.
Similarly, there are 6 flowers remaining for the 3rd space, 5 for
the 4th space, 4 for the 5th space, 3 for the 6th space, 2 for the
7th space, and 1 for the 8th space. Because the question asks for
different orders, the order matters, so multiply the numbers in
the spaces without doing any division to get
. The correct answer is

31. D

The question asks for what value an expression approaches as x
approaches 0. Plug in a value close to 0. Let x = 0.1. In a
calculator, compute

to get 2.098. Eliminate (A), (B),

and (C) as they are not close to 2.098. To determine whether
the answer could be (E), plug in a value for x that is slightly less
than 0. Let x = −0.1. In a calculator, compute

to get

1.898. This is still close to 2, so the function does approach 2 as
x approaches 0. As an alternative, graph

on a graphing

calculator. The graph is a mostly smooth graph with a hole at
(0, 2). Using either method, the correct answer is (D).
32. E

The question asks for f(1). Replace x with 1 in f(x) to get f(1) = |7

− 5(1)| = 2. Go through the choices and find the one equal to 2.
In (A), f(−1) = |7 − 5(−1)| = 12. Eliminate (A). In (B), f(0) = |7 −
5(0)| = 7. Eliminate (B). In (C),

. Eliminate

(C). In (D), f(2) = |7 − 5(2)| = 3. Eliminate (D). In (E),
. Keep (E). The correct answer is (E).
33. B

The question asks for the period of a tangent graph. The period
of a tangent graph in the form y = Atan (Bx + C), the period is
equal to the expression
period is

. In this equation B = 2π, so the
. Alternatively, graph the equation

and find the distance between zeroes. This distance is equal to
the period. The correct answer is (B).
34. D

The question asks for how long it would take Katherine to get
from Point X to Point Y along Maple Street and Elm Street.
Because this is a rate question, use the formula d = rt, with d
representing distance, r representing rate, and t representing
time. According to the question, the rate is 15 kilometers per
hour, so substitute r = 15 to get d = 15t. To get the time, find the

distance. According to the question, the straight-line distance
from Point X to Point Y is 40 kilometers. However, the question
also says that she bikes along Maple Street and Elm Street, so
don’t use the straight-line distance for d. Instead use the sum of
the distances from Point X to the intersection and from the
intersection to Point Y for d. The distance from the Point X to
the intersection is 10. To find the distance from intersection to
Point Y, use the Pythagorean Theorem: a2 + b2 = c2. Draw the
straight-line distance from Point X to Point Y and label it 40.
Since 40 is the hypotenuse, plug in a = 10 and c = 40 to get 102
+ b2 = 402. Simplify to get 100 +b2 = 1600. Subtract 100 from
both sides to get b2 = 1500. Take the square root of both sides to
get b ≈ 38.7298. To get the total distance, add this to the
distance from Point X to the intersection to get d = 10 +
38.7298 = 48.7298. Plug this into the equation to get 48.7298 =
15t. Divide both sides by 15 to get t ≈ 3.2487. This is the time in

hours. Because it is between 3 and 4 hours, eliminate (A), (B),
and (C). The remaining two choices only differ by the remainder
in minutes. To get the remainder in minutes, set up the

. Cross multiply to get x ≈ 15

minutes. Therefore, the total time is 3 hours and 15 minutes.
The correct answer is (D).

35. E

The question asks for what could equal g(x). The choices are in
factored form. In the factored form of a polynomial, each factor
is in the form (x −r) where r is one of the roots, so find the
roots. The roots of an equation are the values of x at which the
value of the function is 0. According to the table g(x) = 0, when
x = −2, 1, and 2. Therefore, the factors (x + 2), (x − 1), and (x −
2) must be included in the equation of the function. Eliminate
(A) and (C). Plug the remaining points into the answer choices.
Begin with f(0) = 2. In (B), f(0) = (0 − 2)(0 − 1)(0 + 2)(0 + 3) =
12, so eliminate (B). In (D), f(0) = (0 − 3)(0 − 2)(0 − 1)(0 + 2) =

−12, so eliminate (D). In (E), f(0) = (0 − 2)(0 − 1)
, so keep (E). The correct answer is (E).
36. E

The question asks for what could NOT be a factor of m, an
integer whose only prime factors are 2, 3, 5, and 13. A factor of
m must, by definition, divide m evenly with no remainder. If a
number is to divide m evenly, every factor of the number must
also be a factor of m. Therefore, if a number has any prime
factor other than 2, 3, 5, or 13, the number cannot be a factor of
m. Find the prime factors of each choice. Because the question
says NOT, ignore the NOT. Instead mark a choice with Y if it
could be a factor and N if it could not be. The prime factors of 6
are 2 and 3. Since 2 and 3 are prime factors of m, this could be a
factor, so mark (A) with Y. The prime factors of 9 are 3 and 3.
Since 3 is a prime factor of m, this could be a factor, so mark (B)
with Y. The prime factors of 12 are 2, 2, and 3. Since 2 and 3 are
prime factors of m, this could be a factor, so mark (C) with Y.
The prime factors of 26 are 2 and 13. Since 2 and 13 are prime
factors of m, this could be a factor, so mark (D) with Y. The
prime factors of 35 are 5 and 7. Since 7 is not a prime factor of
m, this cannot be a factor, so mark (E) with N. Four choices are
marked with Y and one choice is marked with N, so select the
choice marked with N. The correct answer is (E).

37. A

The question asks for the value of x. There are values in the
choices, so PITA. Start with (C). If x = 0.328, then cosx ≈ 0.947
and 4sinx ≈ 1.289. These are not equal, so eliminate (C).
Determine whether the answer must be greater or less than
0.328. Use the unit circle. As the angle goes from 0 to

on the

unit circle, then x-coordinate decreases and y-coordinate
increases. The x-coordinate corresponds with cosine, and the y-

coordinate corresponds with sine. Therefore, as the angle
increases, cosine decreases and sine increase. Since cos (0.328)
< 4sin (0.328), the angle must decrease, so that the cosine
increases and the sine decreases. Eliminate (D) and (E). Try (B).
If x = 0.250, then cosx ≈ 0.969 and 4sinx ≈ 0.990. These are
not equal, so eliminate (B). Only one choice remains but try (A)
to be sure. If x = 0.245, the cosx ≈ 0.970 and 4sinx ≈ 0.970.
Although a calculator indicates that the values are not exactly
equal, the answer choices on the test are rounded, so if they are
extremely close, they can be taken to be equal. Therefore, the
correct answer is (A).
38. D

The question asks for the value of g−1(15). The function g−1 is
defined as the inverse of g. For inverse functions, if g(x) = y,
then g−1(y) = x. Therefore, y = 15 and the question asks for the
value of x. Since the question asks for the value of x and there
are numbers in the choices, PITA. Plug the choices in for x and
eliminate any for which f(x) is not 15. Start with (C). If x = 3.17,
then g(x) = g(3.17) ≈ 11.944. This is not 15, so eliminate (C). The
answer must be greater, so eliminate (A) and (B), as well. Try
(D). If x = 5, then g(x) = g(5) = 15. This matches the target, so
the correct answer is (D).

39. C

The question asks for the 11th term of the Triangular Number
Sequence. To find a term in a recursive sequence, start at the
beginning. The definition states that T1 = 1 and that Tn = Tn− 1 +
n for n > 1. Therefore, T2 = T1 + 2 = 1 + 2 = 3. Continue to T11.
T3 = T2 + 3 = 3 + 3 = 6.
T4 = T3 + 4 = 6 + 4 = 10.
T5 = T4 + 5 = 10 + 5 = 15.

T6 = T5 + 6 = 15 + 6 = 21.
T7 = T6 + 7 = 21 + 7 = 28.
T8 = T7 + 8 = 28 + 8 = 36.
T9 = T7 + 9 = 36 + 9 = 45.
T10 = T9 + 10 = 45 + 10 = 55.
T11 = T10 + 11 = 55 + 11 = 66.
The correct answer is (C).
40. E

The question asks which statements are true about the function
f(x) = x3 + x2 − 16x + 12 are true. If a graphing calculator is
available, graph the function. The graph crosses the x-axis three
times, so statement I is true. Also, the graph has a relative
minimum at (2, −8) and has no points for which y < −8 on the
positive side of the x-axis. Therefore, II and III are also true. If
no graphing calculator is available, then factor to determine the
number of solutions. One method is to test factors of 12 for a
solution. The factors of 12 are 1, 2, 3, 4, 6, and 12. Try x = 1.
Since f(1) = 13 + 12 − 16(1) + 12 = −2, x = 1 is not a solution. Try
x = 2. Since f(2) = 23 + 22 − 16(2) + 12 = −8, x = 2 is not a
solution. Since f(3) = 33 + 32 − 16(3) + 12 = 0, x = 3 is a solution
and (x − 3) is one factor. Rewrite x3 +x2 − 16x + 12 in a way that
makes it easy to factor (x − 3). First, rewrite it as x3 − 3x2 + 3x2
+ x2 − 16x + 12, and factor the first two terms to get x2(x − 3) +
4x2 − 16x + 12. Now, rewrite it as x2(x − 3) + 4x2 − 12x + 12x −
16x + 12, and factor 4x2 − 12x to get x2(x − 3) + 4x(x − 3) − 4x +
12. Now factor −4x + 12 to get x2(x − 3) + 4x(x − 3) − 4(x − 3).
Factor (x − 3) to get (x − 3)(x2 + 4x − 4). Now determine the
number of factors of x2 − 4x + 4. To determine the number of
factors of a quadratic in the form ax2 + bx + c, use the
discriminant: b2 − 4ac. If the discriminant is positive, there are
two real solutions. If the discriminant is 0, there is one real

solution. If the discriminant is negative, there are no real
solutions. In the quadratic x2 + 4x − 4, a = 1, b = 4, and c = −4,
so b2 − 4ac = 42 − 4(1)(−4) = 32 > 0. Since the discriminant is
positive, x2 + 4x − 4 has two solutions and (x − 3)(x2 + 4x − 4)
has three solutions. Thus, (I) is true. Eliminate (B) and (D). Test
(II), which says that f(x) ≥ −8, for all x ≥ 0. Set up x3 + x2 − 16x
+ 12 ≥ −8. Get one side equal to 0. Add 8 to both sides to get x3
+ x2 − 16x + 20 ≥ 0. Similarly, factor the polynomial on the
right by testing the factors of 20: 1, 2, 4, 5, 10 and 20. Since 13 +
12 − 16(1) + 20 = 6, x = 1 is not a solution. Since 23 + 22 − 16(2)
+ 20 = 0, x = 2 is a solution, so (x − 2) is a factor. Rewrite x3 +
x2 − 16x + 20 as x3 − 2x2 + 2x2 + x2 − 16x + 20 = x2(x − 2) + 3x2
− 16x + 20 = x2(x − 2) + 3x2 − 6x + 6x − 16x + 20 = x2(x − 2) +
3x(x − 2) − 10x + 20 = x2(x − 2) + 3x(x − 2) − 10(x − 2) = (x − 2)
(x2 + 3x − 10). Factor to get (x − 2)(x − 2)(x + 5) = (x − 2)2(x +
5). Therefore, the expression x3 + x2 − 16x + 20 = 0 when x = 2
and x = −5. The statement only refers to what happens when x ≥
0, so ignore x = −5. Since x3 + x2− 16x + 20 = 0, when x = 2, x3
+ x2 − 16x + 12 = f(x) = −8, when x = 2. Determine what
happens to the left and right of x = 2. If x = 1, then f(1) = 13 + 12
− 16(1) + 12 = −2 ≥ −8. If x = 3, then f(3) = 33 + 32 − 16(3) + 12
= 0 ≥ −8. Therefore f(x) ≥ −8, whenever x ≥ 0, so (II) is true.
Eliminate (A) and (C). Only one choice remains. The correct
answer is (E).
41. D

The question asks which of the following could be a portion of
the graph of gh, the product of the graphs of g and h. Look at
the graphs in pieces. When x is negative, the graphs of g and h
are both negative. Therefore, the product must be positive.
Eliminate (A) and (C), which are positive when x is negative. If
x is positive, g is positive but h is negative, so the product must
be negative. Eliminate (B), which is positive when x is positive.
If x = 0, the f and g are both 0, so the product is 0, and the
graph of the product must go through the origin. Eliminate (E),
which does not go through the origin. The correct answer is (D).

42. D

The question asks what set of real numbers makes the equation
true. Plug In values of y. Try y = 0. If y = 0, then the equation
is true. Eliminate (B) and (C), which indicate that 0
should not be included. Plug in a positive number. Try y = 2. If
y = 2, then the equation

is true. Eliminate (E), which

says that 2 should not be included. Now, try a negative. Try y =
−2 If y = −2, then the equation

is false.

Eliminate (A), which says that −2 should be included. Only one
choice remains. The correct answer is (D).
43. D

The question asks for sinx. The triangle is not necessarily a right
triangle. In non-right triangles, use the law of sines:
. The side opposite the 60° angle is 12 and the side
opposite the x° angle is 5. Plug these into the formula to get
. Cross multiply to get 5sin (60) = 12sin (x). Use
the 30-60-90 right triangle to get sin (60)
. Divide both sides by 12 to get

. Therefor,
. The

correct answer is (D).
44. C

The question asks for the longest segment that can be drawn
between two vertices of the solid. The longest segments are

always the diagonals that go through the center of the solid. To
find the length of one of them, use the three-dimensional
version of the Pythagorean Theorem: a2 + b2 + c2 =d2, where a,
b, and c represent the dimensions of the solid and d represents
the diagonal. Plug in the dimensions to get 62 + 32 + 22 = d2.
Simplify the left side to get 49 = d2. Take the square root of both
sides to get 7 =d. The correct answer is (C).
45. D

The question asks for logn50. The question also uses variables
to represent logn2 and logn5, so put logn50 into those terms.
Notice that 2 and 5 are the prime factors of 50. Since 50 = 2 ×
52, logn50 = logn(2 × 52). A product within a log is equal to the
sum of the logs. Therefore, logn(2 × 52) = logn(2) + logn(52). An
exponent within a log can be moved in front of the log as a
coefficient. Therefore, logn(2) + logn(52) = logn(2) + 2logn(5).
Since logn2 = a, and logn5 =b, logn(2) + 2logn(5) = a + 2b.
Alternatively, plug in n = 10 and use a calculator to get log 2 ≈
0.301, log 5 ≈ 0.699, and log 50 = 1.699. Therefore, a = 0.301, b
= 0.699, and the target is 1.699. Go through the choices and
eliminate any that aren’t 1.699. Choice (A) is 0.301 + 0.699 = 1,
so eliminate (A). Choice (B) is 0.301 + 0.6992 = 0.790, so
eliminate (B). Choice (C) is (0.301)(0.699)2 = 0.147, so
eliminate (C). Choice (D) is 0.301 + 2(0.699) = 1.699, so keep
(D). Choice (E) is 0.301 + 5(0.699) = 3.796, so eliminate (E).
The correct answer is (D).

46. E

The question asks for tanx. There are variables in the choices, so
plug in. Since cosx = a, plug in for x and find a. To make sure

the numbers are easy, plug in x =
to 60°. If x =

, then

tanx. Since x =

radians, which is equivalent
. The question asks for

, tanx = tan

= tan60° =

≈ 1.73. This is the

target. Go through the choices and eliminate any that aren’t
equal to


. Choice (A) is


. This is not




. This is not

eliminate (B). Choice (C) is






This is

, so keep (E).

, so

. This



. This is not

eliminate (D). Choice (E) is


, so


47. A

The question asks for the shift in the graph. There are variables
in the choices, so plug in. Since c is a constant greater than 5,
plug in c = 6. Then, the equation is y = x2 + 4x + 6. To
determine the shift, find the vertex by getting the equation in
vertex form. Complete the square. Set y = 0 to get 0 = x2 + 4x +
6. Subtract 6 to move the constant to the other side to get −6 =
x2 + 4x. Add the square of half the coefficient on x to both sides.
The coefficient is 4, half of 4 is 2, and the square of 2 is 4. Add 4
to both sides to get −2 = x2 + 4x + 4. Factor the right side to get
−2 = (x + 2)2. Add 2 to both sides to get 0 = (x + 2)2 + 2.
Replace 0 with y to get y = (x + 2)2 + 2. Now, the equation is in
the form y = (x − h)2 + k, where (h, k) represents the vertex of
the parabola. Therefore, the vertex is (−2, 2). Since the vertex of
y = x2 is (0, 0), the graph shifts to the left 2. Eliminate (B), (C),
and (E). The graph also shifts up 2 units. Since c = 6, (A) says
that the graph shifts up 6 − 4 = 2 units, while (D) says that the
graph shifts 6 + 4 = 10. Eliminate (D). The correct answer is

48. B

The question is asking for the percent that the side of the base
must be increased. Since the question is asking for a percent of
some unknown value, the hidden plug in approach is the best to
use here. The volume formula for a rectangular pyramid is V =
lwh, where l and w represent the length and width,
respectively, of the base, and h represents the height of the
pyramid. First, plug in numbers just to calculate a starting
volume. Since the base of the pyramid is a square, the length
and width will be the same. Let l = 2, w = 2, and h = 3. Plugging
into the formula yields: V =

(2)(2)(3) = 4. Next, calculate the

increased volume of the pyramid: 28% of 4 = 0.28 × 4 = 1.12.
The volume that is 28% greater is 4 + 1.12 = 5.12. The question
states that the height is increased by 12%, so calculate the new
height: 12% of 3 = 0.12 × 3 = 0.36. The height that is 12%
greater is 3 + 0.36 = 3.36. Finally, plug the new values
calculated for volume and height into the formula to figure out
the new values for length and width, which will both just equal x
since the values are equal: (5.12) =

(x)(x)(3.36). Simplifying

results in 5.12 = 1.12x2. Divide both sides by 1.12 to get
4.571428571 = x2, and x = 2.138089935. The original side
length was 2, and the question is asking for the percent the side
length needs to be increased. Use the percent change formula of




, which is closest to 7%.
The correct answer is (B).
49. D

The question asks which statements must be true, and the
statements involve the dimensions and existence of matrix
products. In order for a matrix product to exist, the number of
columns of the first matrix must equal the number of rows of
the second matrix. For the product XY, Matrix X has b columns
and Matrix Y has b rows. Therefore, the product does exist. The
number of dimensions matches the rows of the first matrix and
the number of columns of the second. Since Matrix X has a
rows and Matrix Y has c columns, the product XY has
dimensions a × c. Therefore, (I) is true and (II) is false.

Eliminate (B), (C), and (E). For the product YX, Matrix Y has c
columns and Matrix X has a rows. Because these are not equal,
the product YX does not exist. Therefore, (III) is true. Eliminate
(A). The correct answer is (D).
50. E

The question asks what could be iz. Plug in for z, which is a
complex number in Quadrant IV. A complex number in
Quadrant IV is in the form a +bi, where a > 0 and b < 0. Let z =
2 − 3i. Multiply both sides by i to get iz = 2i − 3i2. Since i2 = −1,
iz = 2i − 3(−1) = 3 + 2i. Since both a and b are positive, iz must
be in Quadrant I, and therefore at point E. The correct answer is

When you take the real exam, the proctors will collect your test booklet
and bubble sheet and send your bubble sheet to a processing center
where a computer looks at the pattern of filled-in ovals on your bubble
sheet and gives you a score. We couldn’t include even a small computer
with this book, so we are providing this more primitive way of scoring
your exam.

Determining Your Score

Using the answer key, determine how many questions
you got right and how many you got wrong on the test.
Remember: Questions that you do not answer don’t
count as either right answers or wrong answers.


List the number of right answers here.
(A) ______


List the number of wrong answers here. Now divide that
number by 4. (Use a calculator if you’re feeling
particularly lazy.)
(B) ______ ÷ 4 = (C) ______


Subtract the number of wrong answers divided by 4 from
the number of correct answers. Round this score to the
nearest whole number. This is your raw score.
(A) ______− (C) ______= ______


To determine your real score, take the number from Step
4 and look it up in the left column of the Score

Conversion Table on the next page; the corresponding
score on the right is your score on the exam.

Click here to download and print the PDF version of this table.

Part III
Content Review

Plane and Solid Geometry
Coordinate Geometry
Statistics and Sets

Chapter 5
On the SAT Subject Test in Math 2 you will often be asked to solve for a
variable. While you likely think of yourself as an algebra whiz, ETS has
many tricks up its sleeve to trip you up. Fear not! We’ll show you some
tricks and techniques to avoid falling for ETS’s traps. In addition, we’ll
cover solving for x, inequalities, factoring, simultaneous equations,
quadratics, and more!
(You may be wondering why algebra comes before “fundamentals.” As
you will see in the next chapter, many of our techniques which simplify
algebra problems will also simplify problems about fundamental math
concepts. Trust us!)

Algebra questions will make up approximately 20 percent of the SAT
Subject Test in Math 2. Many of these questions can be best answered
using the algebra rules reviewed in this chapter. Others are best
approached using some of the test-taking techniques discussed in
Chapter 2.

Here are some algebraic terms that will appear on the SAT Subject Test in
Math 2. Make sure you’re familiar with them. If the meaning of any of
these vocabulary words keeps slipping your mind, add those words to
your flash cards.

An unknown quantity in an equation represented by a
letter (usually from the end of the alphabet), for
example, x, y, or z.


An unchanging numerical quantity—either a number
or a letter that represents a number (usually from the
beginning of the alphabet), for example, 5, 7.31, a, b, or


An algebraic unit consisting of constants and variables
including any operation other than addition and
subtraction such as 5x or 9x2.


In a term, the constant before the variable. In ax2, a is
the coefficient. In 7x, 7 is the coefficient.


An algebraic expression consisting of more than one
term joined by addition or subtraction. For example,
x2 − 3x2 + 4x − 5 is a polynomial with four terms.


A polynomial with exactly two terms, such as (x − 5).


A quadratic expression is a polynomial with one
variable whose largest exponent is a 2, for example,
x2 − 5x + 6 or x2+ 4.


A root of a polynomial is a value of the variable that
makes the polynomial equal to zero. More generally,
the roots of an equation are the values that make the
equation true. Roots are also known as zeros, solutions,
and x-intercepts.


The greatest exponent on a variable in a polynomial or
function. For example, f(x) = 3x4 + 3x − 2 is a function
of the fourth degree.

Some questions on the SAT Subject Test in Math 2 will require you to
solve a simple algebraic equation. These questions often present you with
an algebraic equation hidden in a word problem. Setting up an equation
from the information in the problem is the first step to finding a solution
and is the step where careless mistakes are often made. The translation
chart on this page is very useful for setting up equations from
information given in English.
An algebraic equation is an equation that contains at least one unknown
—a variable. “Solving” for an unknown means figuring out its value.
Generally, the way to solve for an unknown is to isolate the variable—that
is, manipulate the equation until the unknown is alone on one side of the
equal sign. Whatever’s on the other side of the equal sign is the value of
the unknown. Take a look at this example.
5(3x3 − 16) − 22 = 18
In this equation, x is the unknown. To solve for x, you need to get x alone.
You isolate x by undoing everything that’s being done to x in the
equation. If x is being squared, you need to take a square root; if x is
being multiplied by 3, you need to divide by 3; if x is being decreased by
4, you need to add 4, and so on. The trick is to do these things in the right
order. Basically, you should follow PEMDAS in reverse. Start by undoing

addition and subtraction, then multiplication and division, then
exponents and roots, and, last, what’s in parentheses.
The other thing to remember is that any time you do something to one
side of an equation, you’ve got to do it to the other side also. Otherwise
you’d be changing the equation, and you’re trying to rearrange it, not
change it. In this example, you’d start by undoing the subtraction.

Then undo the multiplication by 5, saving what’s in the parentheses for

Once you’ve gotten down to what’s in the parentheses, follow PEMDAS in
reverse again—first the subtraction, then the multiplication, and the
exponent last.

At this point, you’ve solved the equation. You have found that the value of
x must be 2. Another way of saying this is that 2 is the root of the
equation 5(3x3 − 16) − 22 = 18. Equations containing exponents may
have more than one root.
Equations which have more than one root are common when you have an

equation with an even degree. For example:
3x4 = 48
First, divide both sides by 3:
x4 = 16
Then take the fourth root of each side. But, because you’re taking an even
root of the equation, you may have a positive OR a negative value for x:

Solving Equations with Radicals
To solve equations with radicals, you work the equation the same way you
work other equations. In order to eliminate the radical, you take both
sides to the power of that radical. For example:

Start by subtracting 2 from both sides, then divide both sides by 3:

Next, square both sides, then divide by 4:
4x = 4; x = 1
Note that you do not get both a positive and negative root when you are
working with radicals. Radical equations will have one solution.

Solving Equations with Absolute Value
The rules for solving equations with absolute value are the same. The only
difference is that, because what’s inside the absolute value signs can be
positive or negative, you’re solving for two different results.

Let’s look at an example:
20. |x −2| = 17
Now we know that either (x − 2) is a negative number or a non-negative
number. When a number is negative, the absolute value makes it the
inverse, or multiplies it by −1 to yield a positive result. If the number is
positive, it remains the same after being sent through the absolute value
machine. So when we remove the absolute value bars, we’re left with two
different equations:
x − 2 = 17


x − 2 = −17

Now simply solve both equations:

And that’s all there is to it!

Practice solving equations in the following examples. Remember that
some equations may have more than one root. The answers can be found
in Part IV.
1. If

, then x =

2. If n2 = 5n, then n =
3. If

4. If

, then a =

, then s =

5. If

, then x =

6. If |2m + 5|=23, then m=
7. If

, then r =

8. If

, then x =

9. If

and y ≠ 0, then y=

When manipulating algebraic equations, you’ll need to use the tools of
factoring and distributing. These are simply ways of rearranging
equations to make them easier to work with.

Factoring simply means finding some factor that is in every term of an
expression and “pulling it out.” By “pulling it out,” we mean dividing each
individual term by that factor, and then placing the whole expression in
parentheses with that factor on the outside. Here’s an example:
x3 − 5x2 + 6x = 0
On the left side of this equation, every term contains at least one x—that
is, x is a factor of every term in the expression. That means you can factor
out an x:
x3 − 5x2 + 6x = 0
x(x2 − 5x + 6) = 0

The new expression has exactly the same value as the old one; it’s just
written differently, in a way that might make your calculations easier.
Numbers as well as variables can be factored out, as seen in the example
11x2 + 88x + 176 = 0
This equation is, at first glance, a bit of a headache. It’d be nice to get rid
of that coefficient in front of the x2 term. In a case like this, check the
other terms and see if they share a factor. In fact, in this equation, every
term on the left side is a multiple of 11. Because 11 is a factor of each term,
you can pull it out:
11x2 + 88x +176 = 0
11(x2 + 8x +16) = 0
x2 + 8x +16 = 0
(x + 4)2 = 0
x = −4
As you can see, factoring can make an equation easier to solve.

Distributing is factoring in reverse. When an entire expression in
parentheses is being multiplied by some factor, you can “distribute” the
factor into each term, and get rid of the parentheses. For example:
3x(4 + 2x) = 6x2 + 36
On the left side of this equation the parentheses make it difficult to
combine terms and simplify the equation. You can get rid of the
parentheses by distributing.

And suddenly, the equation is much easier to solve.

Practice a little factoring and distributing in the following examples, and
keep an eye out for equations that could be simplified by this kind of
rearrangement. The answers can be found in Part IV.
3. If (11x)(50) + (50x)(29) = 4,000, then x =
(A) 2,000
(B) 200
5. If ab ≠ 0;
(A) −3
(B) −2
(C) 0
(D) 1
(E) 3
22. If x ≠ −1,

(A) 4x2
(B) x2
(C) 4x
(D) x
(E) 4
30. If x ≠ 0, −1, then (x5 + 2x4 +x3)−1 =
(B) −x3(x+1)2

Plugging In is a technique for short-cutting algebra questions. It works on
a certain class of algebra questions in which relationships are defined, but
no real numbers are introduced. For example:

1. The use of a neighborhood car wash costs n dollars for a
membership and p cents for each wash. If a membership
includes a bonus of 4 free washes, which of the following
reflects the cost, in dollars, of getting a membership at
the car wash and washing a car q times, if q is greater
than 4 ?
(A) 100n + pq − 4p
(B) n + 100pq − 25p

(C) n + pq −
(D) n +


(E) n +


Here’s How to Crack It
In this problem you see a lot of variables in the question and in the
answer choices. That’s a big clue!
When you see variables in the answer choices, PLUG IN!
Let’s try Plugging In with this problem. We’ll start with n, the
membership fee.
Plug In an easy number like 3, so that a membership costs $3.00.
Then, Plug In a number for p, the charge per wash. Since this number is
in cents, and we’ll need to convert it to dollars in the answers, choose a
number that can be converted easily to dollars, like 200. Let’s make p =
200, so a wash costs $2.00.
To Number or Not to Number?
Let’s say you walk into a candy store. The store is selling certain
pieces of candy for 5 cents and 10 cents each. You want to get 3
pieces of the 5 cent candy and 6 pieces of the 10 cent candy. You give
the cashier a $5 bill. What’s your change?
Ok, now let’s say you walk into a candy store run by ETS. This store
is selling certain pieces of candy for x cents and y cents each. You
want to get m pieces of the x cent candy and n pieces of the y cent
candy. You give the cashier a $z bill. What’s your change?
Which problem would be easier to solve? The one with the numbers!

Numbers make everything easier. So why bother with variables when
you don’t have to?
Last, let’s say that q, the number of washes, is 5. That’s as easy as it gets.
With 4 free washes, you’re paying for only 1.
Then, just work out the answer to the question using your numbers. How
much does it cost for a membership and 5 washes? Well, that’s $3.00 for
a membership, 4 washes free, and 1 wash for $2.00. The total is $5.00.
That means that if you plug your numbers into the answer choices, the
right answer should give you 5. We call that your target number—the
number you are looking for in the answer choices. Put a double circle
around your target number, so that it stands out from all the other
numbers you’ve written down. It looks like a bull’s-eye that you’re trying
to hit:

When you plug n = 3, p = 200, and q = 5 into the answer choices, the only
answer choice that gives you 5 is (D). That means you’ve hit your target
number, and you’re done.
A Big Clue
There will be times when ETS will give you
questions that include variables and the phrase
“in terms of” (for example, “in terms of x”). This
is a big clue that you can Plug In. Cross off the
phrase “in terms of x,” because you don’t need
it to solve the problem.

Let’s look at a problem without variables:

3. The size of an art collection is tripled, and then 70
percent of the collection is sold. Acquisitions then
increase the size of the collection by 10 percent. The size
of the art collection is then what percent of its size
before these three changes?
(A) 240%
(B) 210%
(C) 111%
(D) 99%
(E) 21%
Here’s How to Crack It
Here’s another question in which you aren’t given numbers. In this case,
you don’t know the original size of the art collection. Instead of variables,
though, the question and answers contain percents. This is another sign
that you can Plug In whatever numbers you like. Because you’re working
with percentages, 100 is a good number to Plug In—it’ll make your math
You start with a collection of 100 items. It’s tripled, meaning it increases
to 300. Then it’s decreased by 70%. That’s a decrease of 210, so the
collection’s size decreases to 90. Then, finally, it increases by 10%. That’s
an increase of 9, for a final collection size of 99. Since the collection began
at 100, it’s now at 99% of its original size. The answer is (D). It doesn’t
matter what number you choose for the original size of the collection—
you’ll always get the right answer. The trick to choosing numbers is
picking ones that make your math easier.

Take a look at one last (challenging) problem:

50. If x dollars are invested at y percent annual interest, and
0 < x < 10,000 and y > 0, then in how many years will
the value of the investment equal $10,000?

(A) logx + log (1+0.01y)

Here’s How to Crack It
The answer choices with logarithms make this problem look extra
intimidating. This problem deals with repeated percent change, but the
variables in the question mean you can Plug In numbers to make life
easier. If you make y = 100 then each year the value of the investment
doubles. If you make x = 2,500, then after one year the investment is
worth 5,000 and after two years the investment is worth 10,000. 2 is your
target; circle it. Now you can plug in 2,500 for x and 100 for y in each
answer, eliminating answers that do not equal 2. The only answer that
equals 2 is (E).

Remember to Plug In!
1. See variables in the question and answers?
You can Plug In!
2. Assign a numerical value to each variable.
Pick numbers that make the math easy.
3. Work the problem until you get a numerical
answer. Circle it twice; that’s the target.
4. Plug In your numbers to each answer
choice, eliminating every answer which doesn’t
match your target. Always check every answer
choice with variables!

The idea behind Plugging In is that if these relationships are true, then it
doesn’t matter what numbers you put into the question; you’ll always
arrive at the same answer choice. So the easiest way to get through the
question is to Plug In easy numbers, follow them through the question,
and see which answer choice they lead you to.
Occasionally, more than one answer choice will produce the correct
answer. This often occurs when the question is looking for an exception,
or something that “must be true.” When that happens, eliminate the
answer choices that didn’t work out, and Plug In some different kinds of
numbers. Some numbers you might try are odd and even integers,
positive and negative numbers, fractions, 0, 1 or −1, and really big or
really small numbers, like 1,000 or −1,000. The new numbers will
produce a new target number. Use this new target number to eliminate
the remaining incorrect answer choices. You will rarely have to Plug In
more than two sets of numbers.
When using Plugging In, keep a few simple rules in mind:
• Avoid Plugging In 1 or 0, which often makes more than one
answer choice produce the same number. For the same reason,
avoid Plugging In numbers that appear in the answer choices—
they’re more likely to cause several answer choices to produce
your target number.
• Plug In numbers that make your math easy—2, 3, and 5 are good
choices in ordinary algebra. Multiples of 100 are good in
percentage questions, and multiples of 60 are good in questions
dealing with seconds, minutes, and hours.

Plugging In can be an incredibly useful technique. By Plugging In
numbers, you’re checking your math as you do the problem. When you
use algebra, it takes an extra step to check your work with numbers. Also,
there are fewer chances to mess up when you Plug In. And you can Plug
In even when you don’t know how to set up an algebraic equation.

Plugging In is often safer because ETS designs the answer choices so that,
if you mess up the algebra, your result will be one of the wrong answers.
When your answer matches one of the choices, you think it must be right.
Very tempting. Furthermore, all of the answer choices look very similar,
algebraically. This is how ETS camouflages correct answers. But when
you Plug In, the answers often look very different. Often you’ll be able to
approximate to eliminate numbers that are obviously too big or too small,
without doing a lot of calculation, and that will save you lots of time!
It is strongly recommended that you become comfortable with Plugging
In. As you will see throughout this book, Plugging In often turns a
challenging problem into a piece of cake. Even